google地图、高德地图基于基站定位位置纠偏

拼搏现实的明天。 2023-10-17 08:04 78阅读 0赞

GPS纠偏[算法][Link 1]，适用于google,高德体系的地图,精确度还比较高。我试了一下比高德本身的纠偏还精确点。

/**
     * gps纠偏算法，适用于google,高德体系的地图
     * @author Administrator
     */
    public class GpsCorrect {
    	final static double pi = 3.14159265358979324;
    	final static double a = 6378245.0;
    	final static double ee = 0.00669342162296594323;
    
    	public static void transform(double wgLat, double wgLon, double[] latlng) {
    		if (outOfChina(wgLat, wgLon)) {
    			latlng[0] = wgLat;
    			latlng[1] = wgLon;
    			return;
    		}
    		double dLat = transformLat(wgLon - 105.0, wgLat - 35.0);
    		double dLon = transformLon(wgLon - 105.0, wgLat - 35.0);
    		double radLat = wgLat / 180.0 * pi;
    		double magic = Math.sin(radLat);
    		magic = 1 - ee * magic * magic;
    		double sqrtMagic = Math.sqrt(magic);
    		dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * pi);
    		dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * pi);
    		latlng[0] = wgLat + dLat;
    		latlng[1] = wgLon + dLon;
    	}
    
    	private static boolean outOfChina(double lat, double lon) {
    		if (lon < 72.004 || lon > 137.8347)
    			return true;
    		if (lat < 0.8293 || lat > 55.8271)
    			return true;
    		return false;
    	}
    
    	private static double transformLat(double x, double y) {
    		double ret = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * Math.sqrt(Math.abs(x));
    		ret += (20.0 * Math.sin(6.0 * x * pi) + 20.0 * Math.sin(2.0 * x * pi)) * 2.0 / 3.0;
    		ret += (20.0 * Math.sin(y * pi) + 40.0 * Math.sin(y / 3.0 * pi)) * 2.0 / 3.0;
    		ret += (160.0 * Math.sin(y / 12.0 * pi) + 320 * Math.sin(y * pi / 30.0)) * 2.0 / 3.0;
    		return ret;
    	}
    
    	private static double transformLon(double x, double y) {
    		double ret = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * Math.sqrt(Math.abs(x));
    		ret += (20.0 * Math.sin(6.0 * x * pi) + 20.0 * Math.sin(2.0 * x * pi)) * 2.0 / 3.0;
    		ret += (20.0 * Math.sin(x * pi) + 40.0 * Math.sin(x / 3.0 * pi)) * 2.0 / 3.0;
    		ret += (150.0 * Math.sin(x / 12.0 * pi) + 300.0 * Math.sin(x / 30.0 * pi)) * 2.0 / 3.0;
    		return ret;
    	}
    }

[Link 1]: http://lib.csdn.net/base/31